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I accidentally typed

cd /tmp | tail

which is of course not what I meant. But to my surprise, I then did not change directory. Why is this? For comparison

cd /tmp > /tmp/foo

does result in a change of directory, so it's not the output redirection per sé. What's going on here?

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  • The pipe spawns a subshell that exits when tail exits. I would have expected only tail to be in the subshell, but I guess the cd is as well.
    – jordanm
    Jun 22, 2018 at 16:21
  • 1
    But cd /tmp | pwd exhibits the same behavior -- pwd shows an unchanged working directory.
    – DopeGhoti
    Jun 22, 2018 at 16:23
  • Each part of the pipeline is a separate process (maybe except last command).
    – ctrl-alt-delor
    Jun 22, 2018 at 16:56

1 Answer 1

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3

Pipes involve subshells, portions of which are not run in the parent shell and thus cannot affect the working directory of that parent shell process. This is how ZSH behaves:

% cd /tmp
% cd /var/tmp | tail
% pwd
/tmp
% echo foo | cd /var/tmp
(pwd now: /var/tmp)
% cd /tmp
% echo foo | cd /var/tmp | echo bar
bar
% pwd
/tmp
%

Note how the directory only changed when the cd was the last command in the pipeline; this was run in the parent shell proper and thus was able to change the working directory of that process.

A useful use of this feature usually involves an explicit subshell and commands run therein:

dowork | ( cd elsewhere && domorework ) | andyetmore

also be sure to error check the cd call instead of assuming it worked, unless you like rsync output sprayed all over /, or other such hypothetical messes...

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2
  • I understand why tail would need to be in a subshell, but I don't understand by the cd must be in a subshell, for the pipe to work.
    – gerrit
    Jun 22, 2018 at 23:17
  • having the parent process also run at various random points throughout the pipeline and handling I/O and deadlocks correctly would be much more complicated than simply forking off N children with the I/O wired up correctly
    – thrig
    Jun 23, 2018 at 14:33

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